HW3 - Problem 1

Problem 1: Bell States and Partial Measurements

We’ve discussed the Bell pair in class, which (like any state) can be part of an orthonormal basis. Since these vectors are length four, the basis should have four states. The following is called the Bell basis, formed of the four canonical Bell states.

  • \(\left|\Phi^+\right> := \frac{1}{\sqrt{2}}(\left|00\right> + \left|11\right>)\)
  • \(\left|\Phi^-\right> := \frac{1}{\sqrt{2}}(\left|00\right> - \left|11\right>)\)
  • \(\left|\Psi^+\right> := \frac{1}{\sqrt{2}}(\left|01\right> + \left|10\right>)\)
  • \(\left|\Psi^-\right> := \frac{1}{\sqrt{2}}(\left|01\right> - \left|10\right>)\)
  1. Design a two qubit circuit that starts in the state \(\left|00\right>\), and ends with \(\ket{\Psi^-}\).
  2. For the remaining parts for this problem, we will use the following notation.

\(\left|\theta\right> := \cos(\theta)\left|0\right> + \sin(\theta)\left|1\right>.\)

Consider the basis \(\{\left|\pi/6\right>, \left|4\pi/6\right>\}\). Write down \(\left|0\right>\) and \(\left|1\right>\) as a weighted sum of \(\left|\pi/6\right>\) and \(\left|4\pi/6\right>\) in the above basis. Recall that

\(\cos(\pi/6) = \sqrt{3}/2 \;\;\;\; \sin(\pi/6) = 1/2\)

\(\cos(4\pi/6) = -1/2 \;\;\;\;\sin(4\pi/6) = \sqrt{3}/2\)

  1. Suppose Alice has the first qubit and Bob has the second qubit of a \(\left|\Psi^-\right>\) state. If Alice measures her qubit in the standard basis, what are the probabilities of each outcome, and the state of the two qubits after the measurement?
  2. If Alice instead chooses to measure in the \(\{\left|\pi/6\right>, \left|4\pi/6\right>\}\) basis, at are the probabilities of each outcome, and the state of the two qubits after the measurement? The most general way to accomplish this is to rewrite Alice’s qubit in the Bell basis, and see what the amplitudes are in that form.
  3. Verbally describe what happens to the second qubit when the first qubit of a \(\left|\Psi\right>\) state gets measured.